All categories

3 years ago

Find the value of the variable. Leave in simplest radical form.

Mathematics

1 answer:

Mademuasel [1]3 years ago

8 0

If the hypotenuse is equal to 11, then we can plug it into the Pythagorean theorem. So we get a^2+x^2=121. So we take the square root of both sides, sqrt(a^2+x^2)=11. So we take a^2 and x^2 out of the square root. We get a+x=11. So now we have two equations, a+x=11 and a^2+x^2=121. I’ll leave the rest of solving equations to you.

Hint: a=11-x
You can plug that into a^2+x^2=121

You might be interested in

What is the absolute value of –27? A. –27 B. 27

Serga [27]

Absolute value makes it positive
answe is 27
the answer is B

6 0

3 years ago

Read 2 more answers

Just a short answer please help :( the top

Shkiper50 [21]

26/100 = x/82 I think that's what you want.

3 0

3 years ago

Read 2 more answers

Question Help Suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a

koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a value X is given by:

$Z = \frac{X - \mu}{\sigma}$

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So $\mu = 5656, \sigma = 333.3$

(a) What proportion of light bulbs will last more than 6262 hours?

The pvalue of the z-score of $X = 6262$ is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{6262 - 5656}{333.3}$

$Z = 1.82$

$Z = 1.81$ has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

$P = 100% - 96.562% = 3.438%$ of the light bulbs will last more than 6262 hours.

(b) What proportion of light bulbs will last 5252 hours or less?

This is the pvalue of the zscore of $X = 5252$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5252- 5656}{333.3}$

$Z = -1.21$

$Z = -1.21$ has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 hours?

This is the pvalue of the zscore of $X = 6262$ subtracted by the pvalue of the zscore $X = 5858$

For $X = 6262$ , we have that $Z = 1.81$ with a pvalue of .96562.

For X = 5858

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{5858- 5656}{333.3}$

$Z = 0.61$

$Z = 0.61$ has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

$P = .96562 - .72907 = .23655$

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?

This is the pvalue of the zscore of $X = 4646$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{4646- 5656}{333.3}$

$Z = -3.03$

$Z = -3.03$ has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0

3 years ago

What Is the area of the shaded part???

Aloiza [94]

Answer: The area of the shaded region is 27.74 centimeters squared

Step-by-step explanation: The diagram shows a two-in-one figure, a circle inscribed in a rectangle. The shaded region is the part of the rectangle not covered by the circle, hence we would have to subtract the area of the circle from that of the rectangle.

Area of Rectangle = L x W

Area of rectangle = 8 x 7

Area of rectangle = 56

Also,

Area of circle = Pi x r^2

Area of circle = 3.14 x 3^2

Area of circle = 3.14 x 9

Area of circle = 28.26

Area of shaded region is given as area of rectangle minus area of circle

Therefore, shaded region = 56 - 28.26

Area of shaded region = 27.74 centimeters squared

3 0